For concreteness, I will work in the category of smooth manifolds, but my question makes sense in topological and PL category as well. Recall that a manifold $M$ is called open if every connected component of $M$ is non-compact.

Question. Is it true that for every $n$ there exists a compact $n$-dimensional manifold $N^n$ so that every open $n$-dimensional manifold $M^n$ admits an immersion in $N^n$? (In this context an immersion is just a local diffeomorphism.)

I think that the answer is positive and that the manifolds $N$ are connected sums of products of projective spaces of various dimensions.

Edit: Igor noted that real-projective spaces are not enough, one should include complex-projective spaces as factors in the products. (The reason I think that products real and complex projective spaces are the right thing to use is that real and complex projective spaces generate rings of unoriented and oriented cobordisms.)

Some background: As we know very well, not every manifold admits an open embedding in a compact manifold. For instance, the infinite connected sum of 2-dimensional tori does not. However, it is easy to prove that every open surface admits an immersion in ${\mathbb R}P^2$. Whitehead proved that every open oriented 3-dimensional manifold admits an immersion in ${\mathbb R}^3$ (and, hence, to any 3-dimensional manifold). I also convinced myself (although I do not have a complete proof) that every open non-orinentable 3-manifold admits an immersion in ${\mathbb R}P^2\times S^1$. More generally, every open paralelizable $n$-manifold admits an immersion in ${\mathbb R}^n$. This is a special case of the Hirsch-Smale immersion theory, which reduces existence of immersion from an open manifold to a homotopy-theoretic question about maps of tangent bundles.

Doesn't the connected sum of products of projective spaces have zero rational Pontryagin classes? If so, the same would be true for any immersed submanifold of the same dimension.
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Igor BelegradekJun 19 '12 at 23:21

What does the torus trick, as given in Kirby Siebenmann do for you in regards to this question?
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Scott CarterJun 19 '12 at 23:51

@Igor: You are right, I have to include complex-projective spaces as possible factors.
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MishaJun 20 '12 at 3:49

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May be this paper can help you: A. Phillips, "Submersions of open manifolds", Topology 6 (1967), 171-206. The main result is that, for $M$ open and $\dim M \geq \dim N$, there is a submersion $M \to N$ iff there is a bundle epimorphism $TM \to TN$. From this you get the following: every oriented 4-manifold with the homotopy type of a (not necessarily finite) 2-complex can be immersed in $CP^2$.
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Daniele ZuddasJun 20 '12 at 15:22

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Roughly, you are trying to approximate the $n-1$-skeleton of $BO(n)$ by a closed (compact?) $n$-manifold and its canonical map. I think that the finiteness of the skeleton means that there is a best $n$-manifold, so that all open $n$-manifolds that immerse in a closed $n$-manifold immerse in this one. The Wu formula restricts tangent bundles, so we cannot fully approximate the $n-1$-skeleton. Key question: do all restrictions on tangent bundles of closed manifolds apply to open manifolds? (Brown and Peterson may be relevant. They probably built something more relevant than BO(n).)
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Ben WielandJun 20 '12 at 19:24

1 Answer
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This is the special case annunced in the comment. We assume $M$ to be an oriented 4-manifold with the homotopy type of a 2-complex. We prove that $M$ immerses in $\mathbb{C}P^2$.

In case of compact 4-manifolds with boundary I have a more direct and elementary proof in my paper which does not depend on Phillips theorem.

Anyway, a bundle map $TM \to T\mathbb{C}P^2$ (needed to apply the theorem of Phillips) can be constructed in this way. Firstly, endow $M$ with an almost-complex structure (which exists by our assumptions). There is a nowhere vanishing vector field on $M$, hence $TM$ splits as a Whitney sum of complex rank 1 bundles $TM = \xi \oplus \varepsilon^1$ with $\varepsilon^1$ trivial and $\xi$ a pullback of the complex universal bundle. By cellular approximation $\xi$ is a pullback of the canonical bundle $\gamma^1_1$ on $\mathbb{C}P^1$, so $TM$ is the pullback of $\gamma^1_1 \oplus \varepsilon^1$. Now, rank 4 real oriented vector bundles over $\mathbb{C}P^1$ are classified by $\Bbb Z_2$ (via the second Stiefel-Whitney class), hence $T\mathbb{CP}^2_{|CP^1} = \gamma^1_1 \oplus \varepsilon^1$ (as real vector bundles). So $TM$ is a pullback of $T\mathbb{C}P^2$ by a suitable map $M \to \mathbb{C}P^2$ (with values in $\mathbb{C}P^1$).